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How can I say that the two lines are the same:

Line1=(5,3)+t(-4,8)

Line2=(5,3)+t(-8,16)

They both have the same starting point but can I say they have the same direction vector? It seems like to me they contain different points for a given "t".

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It would be better to use different variables for the parameter in the two lines. If Line 2 were (5,3)+s(-8,16) one could follow Rasmus, come out with t=2s and it would be clearer. –  Ross Millikan Jan 3 '12 at 19:07
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The vector form of the equation you are using is $$\tag{1}{\bf r}(t) = \color{maroon}{\bf p} +t\color{darkgreen} {\bf d}$$ where $\color{maroon}{\bf p}$ gives the initial point on the line (the tip of $\color{maroon}{\bf p}$ is the initial point on the line corresponding to $t=0$) and $\color{darkgreen}{\bf d}$ is the direction vector.

To see why (1) actually gives a line: Note that multiplication of $\color{darkgreen}{\bf d}$ by $t$ just extends, shortens, or reflects $\color{darkgreen}{\bf d}$, it does not change its direction (in the diagram $\color{maroon}{t{\bf d}}$ is the dashed, maroon vector).
So, thinking of adding vectors "tip to tail", the tip of the vector $\color{orange}{{\bf p}+t{\bf d}}$ will always fall on a line.

Note that in (1), we could have used any other vector, such as the blue vector below, that is parallel to $\color{darkgreen}{\bf d}$ as the direction vector, and the corresponding equation would give the same line, just traced out in a different way.

enter image description here

If you think of the line as being generated by a moving point (the gray one above) whose position at time $t$ is ${\bf r}(t)$, then in your example, the point is moving twice as fast in your equation for line2 than as in the equation for line1.

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Yes, they are because $(-8,16)$ is a multiple of $(-4,8)$:

$$(-8,16)=2\cdot (-4,8).$$

If a point $(a,b)$ belongs to Line1, say $(a,b)=(5,3)+t(-4,8)$, then it also belongs to Line2 because $(a,b)=(5,3)+\dfrac{t}{2}(-8,18)$.

Similarly, if a point $(a,b)$ belongs to Line2, say $(a,b)=(5,3)+t(-8,16)$, then it also belongs to Line1 because $(a,b)=(5,3)+2t(-4,8)$.

The lenght of the "direction vector" is irrelevant, only its direction is important. This is because your are really only interested in the set of all scalar multiples of the direction vector.

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Yes, they are the same since the direction vectors are parallel (i.e. one is a scalar multiple of the other.) It is true that they contain different points for a given "t", but they still contain all the same points -- for instance, the point of line1 at a certain $t$ corresponds to the point at line2 at time $t/2$.

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